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  1. Sep 20, 2008 #1
    If I build a tower - let's say, on the equator - and every so often, as the tower gets higher, I drop a stone from its top. Presumably, as I get fiurther away from the Earth's surface, each stone will accelerate towards ground at a lesser rate than one dropped when the tower was lower. How high will the tower need to be for a dropped stone to depart from Earth rather than be attracted to it, and what are the physics/mathematics behind the answer?
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  3. Sep 20, 2008 #2

    Doc Al

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    Are you considering the Earth's rotation? If not, then the stone would always fall to the Earth, as gravity always acts. Of course gravity does get weaker as you get further from the Earth, but it doesn't become negative.
  4. Sep 20, 2008 #3
    My tower is static with respect to the Earth, but shares all of Earth's motion.
  5. Sep 20, 2008 #4

    Doc Al

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    Then the higher you go, the greater your speed: [itex]v = \omega R[/itex], where R is the distance from the center of the Earth.

    Also, the less energy you need to escape the Earth's gravity, which is given by [itex]1/2mv^2 = GMm/R[/itex].

    That should allow you to compute the height of the tower, in this very hypothetical thought experiment, that will have the released stone escape into space.
  6. Sep 20, 2008 #5
    Presumably, then, I will eventually reach a height at which I can release a stone and it will neither fall nor rise. If I then throw a stone horizontally east, and another horizontaly west, one will have a velocity less than v, and the other will have a velocity greater than v, where v = omega * R (by the way, how do you write the omega symbol?). Does that mean that one will fall to earth, and the other depart?
  7. Sep 20, 2008 #6
  8. Sep 20, 2008 #7

    Doc Al

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    Yes, but that point is somewhat different (not as high) as the height needed to reach escape velocity. That point is where the gravitational force is just enough to provide the required centripetal force needed to keep the stone in geostationary orbit. (That would be one tall tower. About 5.6 Earth radii tall. :wink:)
    If you threw it fast enough towards the east, it could end up with escape velocity. But in general, giving it a bit higher or lower speed by tossing it one way or the other will give it a different orbit around the Earth's center. (An orbit that may well have it smack into the ground.)

    (I used Latex to write [itex]\omega[/itex]. Click on it to see the code.)
  9. Sep 20, 2008 #8


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    No, actually. Althought that seems like the natural conclusion, what you're getting involved in here is orbital dynamics, and orbtis don't behave the way you might think they would. The rock thrown East would boost into an elliptical orbit with its highest point above the spot from which you threw it, and so that orbit would take a little longer. The other rock (thrown west), would be slowed down, and so would fall into a lower orbit. Because it is a lower robit, it is a faster orbit. Strange as it sounds, throwing East will amke the rock fall behind you (to the West) as you stand there in the tower, and throwing West will make it speed up and go to the East of you. Weird, huh?

    BTW, the height at which you could drop the stone and it would just "hover" is about 22,365 miles up. This is where sattelites in geosynchronous orbit are located, and the rocjk would essentially be in that orbit.
  10. Sep 20, 2008 #9
    Then does that mean that a person driving his Ferrari around the equator going east weighs less than one driving west? Does it mean that if I throw a stone westwards, standing on Earth's surface, and another to the east, one will reach ground before the other?

    I'm sorry to be asking such apparently daft questions, but I'm trying very hard to understand something that I hope my questions will eventually get to.
  11. Sep 20, 2008 #10


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    Maybe you should just skip to the punchline. I'm sure there are people here qualified to answer whatever question is bothering you.

    As far as I know, the answer to both your latest questions is: NO
  12. Sep 21, 2008 #11
    My question was intended as an opener to a problem that I have with the concept of absolute motion. As I understand it, and I understand physics only at a fairly elementary level, absolute motion doesn't exist. If I stand and watch a train go by while someone on it throws up a ball and catches it, I see the ball undergo a parabolic motion and return to the hand of the thrower who has a linear motion. With a small amount of coaching and/or revision (I left school many years ago) I could work out the equations involved. If I were watching from a point on the train there would be only one motion involved - that of the ball - and the equations would be much simpler. The point is that the train - or indeed the observer on the track - can be defined as having any velocity whatsoever so long as the relationship between their velocities is retained, and the equations will always explain why the ball returns to the thrower's hand. There is no absolute velocity, no absolute frame of reference.

    To get back to my problem, the first post in response to my question hit the nail right on the head when it mentioned Earth's rotation. What is Earth's rotation? With respect to what? Similarly, what is the velocity of the top of my tower? If frames of reference are not absolute, can we not use one in which the angular velocity of the Earth, and hence the velocity v of the top of the tower, are zero. In such a frame of reference, there is a gravitational force GMm/R(squared) - as always - but no countervailing force from rotation. If my tower reaches the height of a geo-stationary satellite, the same holds true in this reference frame; there is no rotation. Therefore if I drop a stone, or a dead sheep, or anything else, at that point, it should plummet to Earth.

    Now even I can see that there is a flaw here, the chief observational evidence being that geostationary satellites don't land in my garden. The only conclusion that, in my ignorance, I can draw from this is that some frames of reference - perhaps rotating ones only? - must be considered as "absolute", and this causes me all sorts of other problems.

    I have often seen references to motion "with respect to the fixed stars", but is this any more than a convenience, relying on the fact that fhe "fixed" stars only appear so because of their enormous separation from us, while in fact they are all hurtling around the universe at vast velocities in all sorts of directions? When the phrase "the Earth's rotation" is used, what is the reference frame? Is it, in fact, the non-fixed "fixed" stars?

    Some of the previous posts have immediately raised further questions in my mind, apart from various other ones that were there previously, all related to this subject. Perhaps if someone can clear up the confusion about absolute motion (or frames of reference) these other questions may resolve themselves. If not, perhaps I can raise them later.

    Thanks to everyone who has replied so far.
  13. Sep 21, 2008 #12


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    I've been debating over the best way to approach this question. I thought about the differences between inertial and accelerating reference frames, but that gets pretty complicated pretty quick. So I'm gonna try a different tack.

    When a body rotates, it revolves around its own center of mass. So, you could say that the motion of rotation is relative to the center of mass for that object. To any observer, from any reference frame, the object's surface (or other extremity) is moving along a path that is different from the path described by its center of gravity.

    Does that help?
  14. Sep 21, 2008 #13
    Thank you for the answer, Lurch. I shall be without access to the internet for the next week or so, and I'll respond to it (and any others that may turn up) when I return.
  15. Oct 11, 2008 #14
    Hello Lurch. I'm back, later than expected. You say "When a body rotates", and this gets straight to the point of my confusion/questioning/ignorance - call it what you will. What is meant by "a body rotates"? I gave a previous example of someone tossing a ball on a train, and made the point - obvious to anyone on this forum - that the perceived motion of the ball depends on the motion of the observer, i.e. whether he's on the train, on the track, or anywhere else.

    I am an observer of the Earth who is situated on the Earth. Why, to me, is the Earth rotating? Suppose, by magic, all other heavenly bodies suddenly evaporated - how could anyone then assert that the Earth was rotating? I feel no force that tells me that the Earth is spinning, so why does a geo-stationary satellite?

    I apologise again for questions that may seem naive, but I am trying to understand basic concepts that once seemed obvious but no longer do.
  16. Oct 11, 2008 #15


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    Newtons's 2nd and 3rd laws assert that in a "stationary" reference frame all the laws of physics can be written in this simple way:
    2. Accelerations are caused by forces.
    3. Forces come in pairs: a particle only feels a force if it also exerts a force on another particle.

    To make these laws true, we attribute properties like mass or charge to particles, and claim that those properties produce forces, and amazingly we can do that in a way consistent with the Newton's 2nd and 3rd laws.

    On the rotating Earth, to describe all observed accelerations in a way consistent the 2nd law, we have to postulate a "Coriolis force" which violates the 3rd law. So either Newton's laws are wrong, or we are not in a "stationary" frame. If we postulate that Earth is rotating, we don't need the Coriolis force, and we can describe all observed accelerations using only forces consistent with both Newton's 2nd and 3rd laws.

    Peraire's lectures 13-15:
  17. Oct 11, 2008 #16

    D H

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    With respect to inertial :biggrin: -- and that is exactly the terminology that is used. While position and velocity are not absolute, acceleration and thus rotation is absolute.

    You are right on track. If you think about it for a bit, that rotation is absolute is a direct consequence of acceleration being absolute.

    It used to be exactly that, up until 1998. It was not a convenience; the observations of stars over years formed the basis for defining an inertial reference frame. A lot of painstaking number crunching (thank goodness for graduate students) was needed to remove the proper motion of the stars from the picture. The last such star-catalog based inertial reference frame was the "J2000" frame, based on the Fifth Fundamental Catalogue of Stars (FK5).

    So what happened in 1998? We still use stars -- a special kind of stars called quasars. They are so far away that if they their intrinsic motion, is unobservable. Our best guess as to what constitutes an inertial (non-rotating) reference frame is the Inertial Celestial Reference Frame, which supplanted the FK5-based J2000 frame.

    For more info, I suggest you poke around at the following web sites:

    The International Earth Rotation and Reference Systems Service, www.iers.org. The IERS define and disseminate standards on time, earth rotation, and reference frames. (The three concepts are intimately related.)

    The US Naval Observatory, www.usno.navy.mil. All things related to international standards are the property of the French (or so it seems). If that weren't the case, the USNO would be the ones defining the standards on time, earth rotation, and reference frames. As it is, the USNO is the premier organization that contributes to the IERS.
  18. Oct 12, 2008 #17


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    Ah, well, that is what I was trying to clear up by stating that the motion is relative to the center of mass. You said that resorting to "absolute motion" gets you in trouble, so referencing everything off the center of mass gives you something for the motion to be relative to.

    The only reason you don't feel these forces is because they're too slight, compared to the other forces that you do feel. As you stand on the surface of the Earth, gravity is pulling down, and the ground is pushing up under your feet. Compared to these forces, the centrifugal force generated by Earth's rotation is too small to be "felt." However, if you have sensitive enough equipment, it can be measured. Your weight at the equator will be slightly less than your weight at the polls. This is because at the polls, your body is being acted upon by gravity, almost exclusively. But at the equator, your body feels the downward force of gravity minus the upward lift you get from centrifugal force trying to fling you off the surface. It is this same centrifugal force that keeps a geostationary satellite in orbit.

    Not a bit of it; these are difficult concepts to grasp.
    Last edited: Oct 13, 2008
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